/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 14 A toroid having a cross section ... [FREE SOLUTION] | 91影视

91影视

A toroid having a cross section of rectangular shape is defined by the following surfaces: the cylinders \(\rho=2\) and \(\rho=3 \mathrm{~cm}\), and the planes \(z=1\) and \(z=2.5 \mathrm{~cm}\). The toroid carries a surface current density of \(-50 \mathbf{a}_{z} \mathrm{~A} / \mathrm{m}\) on the surface \(\rho=3 \mathrm{~cm}\). Find \(\mathbf{H}\) at the point \(P(\rho, \phi, z):(a) P_{A}(1.5 \mathrm{~cm}, 0\), \(2 \mathrm{~cm}) ;\left(\right.\) b) \(P_{B}(2.1 \mathrm{~cm}, 0,2 \mathrm{~cm}) ;\) (c) \(P_{C}(2.7 \mathrm{~cm}, \pi / 2,2 \mathrm{~cm}) ;\) (d) \(P_{D}(3.5 \mathrm{~cm},\), \(\pi / 2,2 \mathrm{~cm})\)

Short Answer

Expert verified
Question: Determine the magnetic field intensity (H) at points PA(1.5 cm, 0, 2 cm), PB(2.1 cm, 0, 2 cm), PC(2.7 cm, 蟺/2, 2 cm), and PD(3.5 cm, 蟺/2, 2 cm) for a toroid with a surface current density of -50 饾憥饾懅 A/m on the surface 蟻=3 cm. Answer: The magnetic field intensity (H) at the four points are as follows: - H(PA) = 0 - H(PB) = -47.62 饾憥饾湋 A/m - H(PC) = -55.56 饾憥饾湋 A/m - H(PD) = 0

Step by step solution

01

1. Identify given parameters and the Ampere's Circuital Law formula

The given parameters are: - The toroid has a cross section of rectangular shape and is defined by the cylindrical surfaces at 蟻=2 cm and 蟻=3 cm, and the plane surfaces at z=1 cm and z=2.5 cm. - The toroid has a surface current density of -50 饾憥饾懅 A/m on the surface 蟻=3 cm. - We need to find the magnetic field intensity at the four points: 饾憙饾惔(1.5 cm, 0, 2 cm), 饾憙饾惖(2.1 cm, 0, 2 cm), 饾憙饾惗(2.7 cm, 蟺/2, 2 cm), and 饾憙饾惙(3.5 cm, 蟺/2, 2 cm). According to Ampere's Circuital Law, the line integral of the magnetic field intensity (H) over a closed path, denoted as 鈭潗宦穌饾惪, is equal to the net current enclosed by the path (I), which can be represented as: 鈭潗宦穌饾惪 = 饾惣
02

2. Calculate the magnetic field intensity (H) at point PA

For point PA, we can see that it lies outside the toroid since 蟻<2 cm. Therefore, there is no current enclosed by the closed path. By applying Ampere's Law: 鈭潗宦穌饾惪 = 0 Thus, H(PA) = 0
03

3. Calculate the magnetic field intensity (H) at point PB

For point PB, we can see that it lies inside the toroid with 蟻>2 cm and 蟻<3 cm. Therefore, we only need to consider the current within the closed path up to 蟻=2 cm, since it is the nearest surface with a current. Due to the azimuthal symmetry of the toroid, the magnetic field intensity H can be assumed to have only an 蠒 component (饾惢蠒): 鈭潗幌昫饾惪 = -50 A/m 脳 (2蟺 脳 2 cm) Hence, H蠒 = (-50 A/m) 脳 (2蟺 脳 2 cm) / (2蟺 脳 2.1 cm) = -47.62 饾憥饾湋 A/m Thus, H(PB) = -47.62 饾憥饾湋 A/m
04

4. Calculate the magnetic field intensity (H) at point PC

For point PC, it lies on the surface where the toroid has a surface current density, 蟻=3 cm. So, we need to consider the current enclosed by the entire cross-sectional area of the toroid: 鈭潗幌昫饾惪 = -50 A/m 脳 (2蟺 脳 3 cm) Since the path goes through point PC, it is equal to H蠒 脳 (2蟺 脳 2.7 cm): H蠒 = (-50 A/m) 脳 (2蟺 脳 3 cm) / (2蟺 脳 2.7 cm) = -55.56 饾憥饾湋 A/m Thus, H(PC) = -55.56 饾憥饾湋 A/m
05

5. Calculate the magnetic field intensity (H) at point PD

For point PD, it lies outside the toroid since 蟻>3 cm. Therefore, there is no net enclosed current in the closed path. By applying Ampere's Law: 鈭潗宦穌饾惪 = 0 Thus, H(PD) = 0 Now, we have the magnetic field intensity at all four points: - H(PA) = 0 - H(PB) = -47.62 饾憥饾湋 A/m - H(PC) = -55.56 饾憥饾湋 A/m - H(PD) = 0

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Toroid
A toroid is a fascinating shape often used in physics and engineering, especially in studies involving magnetic fields. Imagine a doughnut-like figure, but instead of delicious flavors, we have wires coiled around it to create a toroid. In more technical terms, a toroid is a toroidal coil or wire wound in a doughnut shape. This configuration is particularly useful for generating magnetic fields. It confines magnetic field lines within the center of its coils, which means electromagnetic fields rarely escape outside. This property is essential for devices like transformers and inductors. In this exercise, the toroid is defined by the circular surfaces at distances
  • Inner radius: \( \rho = 2 \, \mathrm{cm} \)
  • Outer radius: \( \rho = 3 \, \mathrm{cm} \)
Additionally, it has flat surfaces defined by planes at:
  • Lower plane: \( z = 1 \, \mathrm{cm} \)
  • Upper plane: \( z = 2.5 \, \mathrm{cm} \)
These boundaries produce a clear and confined magnetic field structure within the toroidal coil.
Magnetic Field Intensity
Magnetic field intensity, represented by \( \mathbf{H} \), is a measure of the magnetizing force within a material. It plays a crucial role in understanding how magnetic fields behave, especially inside materials. In the context of a toroid, we're interested in how strong the magnetic field inside the toroid is and how it influences points inside and around it. According to Ampere's Circuital Law, the magnetic field can be determined by the current distribution in a closed loop.For instance, at points inside the toroid's windings, magnetic field intensity will be determined by integrating the current density over the loop enclosed by the chosen points. However, outside the toroid, there appears to be no net magnetic field because the fields cancel out. This is why points like \( P_D = (3.5 \, \mathrm{cm}, \, \phi, \, 2 \, \mathrm{cm}) \) show zero magnetic field intensity in this context.
Cylindrical Coordinates
Cylindrical coordinates are a way to extend two-dimensional polar coordinates to three dimensions. This system uses three parameters
  • \( \rho \): the radial distance from the origin,
  • \( \phi \): the angular position around the z-axis, and
  • \( z \): the height above the xy-plane.
This coordinate system matches nicely with the geometry of a toroid, which often wraps around a cylinder. In this exercise, all points are defined using cylindrical coordinates, such as - \( P_A = (1.5 \, \mathrm{cm}, \, 0, \, 2 \, \mathrm{cm}) \) and - \( P_C = (2.7 \, \mathrm{cm}, \, \pi/2, 2 \, \mathrm{cm}) \).These coordinates help us easily navigate the positions of points like where the magnetic field intensity needs to be calculated. Understanding the spatial layout using these parameters allows us to determine if a point lies within or outside the toroid, affecting the net magnetic fields experienced there.
Surface Current Density
Surface current density, denoted as \( \mathbf{K} \), is a measure of the current flowing through a unit area of surface. It is essentially talking about how closely packed current carriers are on a surface. In our toroidal configuration, the surface current density is significant because it influences the magnetic field the toroid generates.For the given toroid, at \( \rho = 3 \, \mathrm{cm} \),we know the surface current density is \( -50 \, \hat{\mathbf{a}}_z \, \mathrm{A/m} \).This negative sign indicates the direction of the current, which is crucial for determining the orientation of the magnetic fields produced. The density is a crucial factor in applying Ampere鈥檚 Circuital Law, used to calculate the magnetic field intensity. When dealing with points such as \( P_B = (2.1 \, \mathrm{cm}, \, 0, \, 2 \, \mathrm{cm}) \), inside the toroid, the current distribution within the closed path contributes to calculating the resulting field magnitude and direction.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Show that the line integral of the vector potential A about any closed path is equal to the magnetic flux enclosed by the path, or \(\oint \mathbf{A} \cdot d \mathbf{L}=\int \mathbf{B} \cdot d \mathbf{S}\).

Show that \(\nabla_{2}\left(1 / R_{12}\right)=-\nabla_{1}\left(1 / R_{12}\right)=\mathbf{R}_{21} / R_{12}^{3}\).

Given \(\mathbf{H}=\left(3 r^{2} / \sin \theta\right) \mathbf{a}_{\theta}+54 r \cos \theta \mathbf{a}_{\phi} \mathrm{A} / \mathrm{m}\) in free space: \((a)\) Find the total current in the \(\mathbf{a}_{\theta}\) direction through the conical surface \(\theta=20^{\circ}, 0 \leq \phi \leq 2 \pi\), \(0 \leq r \leq 5\), by whatever side of Stokes' theorem you like the best. \((b)\) Check the result by using the other side of Stokes' theorem.

Assume that \(\mathbf{A}=50 \rho^{2} \mathbf{a}_{z} \mathrm{~Wb} / \mathrm{m}\) in a certain region of free space. \((a)\) Find \(\mathbf{H}\) and \(\mathbf{B}\). \((b)\) Find \(\mathbf{J} .(c)\) Use \(\mathbf{J}\) to find the total current crossing the surface \(0 \leq \rho \leq 1,0 \leq \phi<2 \pi, z=0 .(d)\) Use the value of \(H_{\phi}\) at \(\rho=1\) to calculate \(\oint \mathbf{H} \cdot d \mathbf{L}\) for \(\rho=1, z=0\)

Assume that there is a region with cylindrical symmetry in which the conductivity is given by \(\sigma=1.5 e^{-150 \rho} \mathrm{kS} / \mathrm{m}\). An electric field of \(30 \mathbf{a}_{z} \mathrm{~V} / \mathrm{m}\) is present. ( \(a\) ) Find \(\mathbf{J}\). \((b)\) Find the total current crossing the surface \(\rho<\rho_{0}\), \(z=0\), all \(\phi\). ( \(c\) ) Make use of Amp猫re's circuital law to find \(\mathbf{H}\).

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.